Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
const int maxn = 7920;
typedef long long ll;
typedef std::vector<ll> vec;
std::set<ll> a0,a1;
vec x0,x1;
int ok[maxn],n;
inline void init(std::set<ll>&s,vec&y){
	y.push_back(*s.begin()); s.erase(s.begin());
	const ll lim = 3e7;
	for(;s.size();){
		auto it = s.upper_bound(lim / *y.rbegin()); --it;
		y.push_back(*it); s.erase(it);
	}
}
int main() {
	std::ios::sync_with_stdio(false),std::cin.tie(0);
	for(int i = 2,c=0;i < maxn;++i)if(!ok[i]){
		(++c >> 1 & 1 ? a0 : a1).insert(i);
		for(int j = i;j < maxn;j += i)ok[j] = 1;
	}
	init(a0,x0), init(a1,x1);
	std::cin >> n;
	for(int i = 1;i <= n;++i) for(int j = 1;j <= n;++j) {
		if(i + j & 1) {
			std::cout << x0[i + j - 3 >> 1] * x1[i + n - j - 2 >> 1] * x0[i + j - 1 >> 1] * x1[i + n - j >> 1] + 1;
		} else {
			std::cout << x0[i + j - 2 >> 1] * x1[i + n - j - 1 >> 1];
		}
		std::cout << " \n"[j == n];
	}
}